My Math Forum Evaluating a limit

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 August 15th, 2018, 07:32 PM #1 Senior Member     Joined: Jan 2012 Posts: 123 Thanks: 2 Evaluating a limit Hi, Please evaluate the following limit: $\displaystyle \lim_{x->0}\left ( \tfrac{1}{x^{2}}-\cot^{2}x \right )$ Thanks. Last edited by skipjack; August 15th, 2018 at 08:41 PM.
 August 15th, 2018, 08:03 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 555 Thanks: 319 Math Focus: Dynamical systems, analytic function theory, numerics The limit is 2/3.
August 15th, 2018, 08:27 PM   #3
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Quote:
 Originally Posted by SDK The limit is 2/3.

Last edited by skipjack; August 15th, 2018 at 09:29 PM.

 August 15th, 2018, 08:44 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,598 Thanks: 2583 Math Focus: Mainly analysis and algebra Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}-\cot^2{x}\right) = L$$ setting $2t=x$ we have that \begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}-\cot^2{2t}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{(\cos^2{t}-\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\ 4L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ 3L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) - \lim_{t \to 0} \left(\frac1{t^2}-\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} - \frac1{t^2}+\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\ &=2 \\ L&=\frac23 \end{align} Thanks to this for inspiration. Thanks from happy21 and topsquark
August 15th, 2018, 09:21 PM   #5
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 Originally Posted by v8archie Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}-\cot^2{x}\right) = L$$ setting $2t=x$ we have that \begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}-\cot^2{2t}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{(\cos^2{t}-\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\ 4L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ 3L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) - \lim_{t \to 0} \left(\frac1{t^2}-\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} - \frac1{t^2}+\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\ &=2 \\ L&=\frac23 \end{align} Thanks to this for inspiration.
Brilliant! This method is very nice though uncommon, I think!

Thx.

 August 15th, 2018, 10:06 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,262 Thanks: 1958 \displaystyle \begin{align*}\cot^2\!x &= 1/(\sin^2\!x) - 1 \\ &= 1/(x^2(1 - x^2/3 \,+\; ...)) - 1 \\ &= (1/x^2)(1 + x^2/3 \,+\; ...) - 1 \\ &= 1/x^2 - 2/3 \,+\; ...\end{align*} so the required limit is 2/3. Thanks from happy21
August 16th, 2018, 06:06 AM   #7
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Quote:
 Originally Posted by happy21 Can you please explain?
If you show me some work I'll show you some work.

August 16th, 2018, 10:59 AM   #8
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Quote:
 Originally Posted by SDK If you show me some work I'll show you some work.
Yeah I did from the methods depicted above.

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